The curve y = f(x) is such that the tangent to the curve at every point (x, y) has a Y-axis intercept c, given by c = -y. Then f(x) is proportional to A. x-1 B. x2 C. x3 D. x4

The correct answer is $\boxed{\text{A}. x^{-1}}$.

The slope of the tangent line to the curve at any point $(x, y)$ is $y'(x)$. The $y$-axis intercept of the tangent line is $y'(x)x$. We are given that $y'(x)x = -y$. This means that $y'(x) = -\frac{y}{x}$.

The derivative of $x^{-1}$ is $-\frac{1}{x^2}$. Therefore, $y'(x) = -\frac{y}{x}$ is equivalent to $y = -\frac{1}{x^2}x$. This can be rewritten as $y = -\frac{1}{x}$.

Therefore, $f(x)$ is proportional to $x^{-1}$.

Here is a brief explanation of each option:

• Option A: $x^{-1}$. This is the correct answer. The derivative of $x^{-1}$ is $-\frac{1}{x^2}$. Therefore, $y'(x) = -\frac{y}{x}$ is equivalent to $y = -\frac{1}{x}$. This can be rewritten as $y = -\frac{1}{x^2}x$.
• Option B: $x^2$. This is not the correct answer. The derivative of $x^2$ is $2x$. Therefore, $y'(x) = -\frac{y}{x}$ is not equivalent to $y = -2x$.
• Option C: $x^3$. This is not the correct answer. The derivative of $x^3$ is $3x^2$. Therefore, $y'(x) = -\frac{y}{x}$ is not equivalent to $y = -3x^2$.
• Option D: $x^4$. This is not the correct answer. The derivative of $x^4$ is $4x^3$. Therefore, $y'(x) = -\frac{y}{x}$ is not equivalent to $y = -4x^3$.